I need to get the last record of each month/year for each id.
My table captures daily, for each id, an order value which is cumulative. So, I need that at the end I only have the last record of the month for each id.
I believe without something simple, but with the examples found I could not replicate for my case.
Here is an example of my input data and the expected result: db_fiddle.
My attempt doesn't include grouping by month and year:
select ar.id, ar.value, ar.aquisition_date
from table_views ar
inner join (
select id, max(aquisition_date) as last_aquisition_date_month
from table_views
group by id
)ld
on ar.id = ld.id and ar.aquisition_date = ld.last_aquisition_date_month
You could do this:
with tn as (
select
*,
row_number() over (partition by id, date_trunc('month', aquisition_date) order by aquisition_date desc) as rn
from table_views
)
select * from tn where rn = 1
The tn cte adds a row number that counts incrementally in descending order of date, for each month/id.. Then you take only those with rn=1, which is the last aquisition_date of any given month, for each id
Related
The output should be count of max items sold in a date.
This is bigquery table:
item,date
apple,1-1-2020
apple,1-1-2020
pear,1-1-2020
pear,1-1-2020
pear,1-2-2020
pear,1-2-2020
pear,1-2-2020
orange,1-2-2020
Expected output:
item,date
apple,1-1-2020
pear,1-1-2020
pear,1-2-2020
Consider below approach
select item, date, count(1) sales
from `project.dataset.table`
group by item, date
qualify rank() over(partition by date order by sales desc) = 1
When applied to sample data in your question - output is
If for some reason, you don't want to have sales column in your output - use below
select item, date
from `project.dataset.table`
group by item, date
qualify rank() over(partition by date order by count(1) desc) = 1
if applied to sample data in your question - output is
The following query should do it:
SELECT
item,
sale_date,
FROM (
SELECT
sample.*,
COUNT(item) AS item_count
FROM
sample
GROUP BY
sample.sale_date,
item )
# Here you need to use a WHERE (or HAVING, or GROUP BY) in order to be able to use QUALIFY
WHERE sale_date IS NOT NULL
QUALIFY RANK() OVER(PARTITION BY sale_date ORDER BY item_count DESC) = 1
I have a peculiar problem at hand. I need to rank in the following manner:
Each ID gets a new rank.
rank #1 is assigned to the ID with the lowest date. However, the subsequent dates for that particular ID can be higher but they will get the incremental rank w.r.t other IDs.
(E.g. ADF32 series will be considered to be ranked first as it had the lowest date, although it ends with dates 09-Nov, and RT659 starts with 13-Aug it will be ranked subsequently)
For a particular ID, if the days are consecutive then ranks are same, else they add by 1.
For a particular ID, ranks are given in date ASC.
How to formulate a query?
You need two steps:
select
id_col
,dt_col
,dense_rank()
over (order by min_dt, id_col, dt_col - rnk) as part_col
from
(
select
id_col
,dt_col
,min(dt_col)
over (partition by id_col) as min_dt
,rank()
over (partition by id_col
order by dt_col) as rnk
from tab
) as dt
dt_col - rnk caluclates the same result for consecutives dates -> same rank
Try datediff on lead/lag and then perform partitioned ranking
select t.ID_COL,t.dt_col,
rank() over(partition by t.ID_COL, t.date_diff order by t.dt_col desc) as rankk
from ( SELECT ID_COL,dt_col,
DATEDIFF(day, Lag(dt_col, 1) OVER(ORDER BY dt_col),dt_col) as date_diff FROM table1 ) t
One way to think about this problem is "when to add 1 to the rank". Well, that occurs when the previous value on a row with the same id_col differs by more than one day. Or when the row is the earliest day for an id.
This turns the problem into a cumulative sum:
select t.*,
sum(case when prev_dt_col = dt_col - 1 then 0 else 1
end) over
(order by min_dt_col, id_col, dt_col) as ranking
from (select t.*,
lag(dt_col) over (partition by id_col order by dt_col) as prev_dt_col,
min(dt_col) over (partition by id_col) as min_dt_col
from t
) t;
I've a table that has this information:
And need to get the following information:
If the country of the same person name (in this case Artur) is different, then I need to sum the two values of quantity from the max date (in this case 04/10) and return both person (Artur) and the qty (15k)
If the country of the same person name (in this case Joseph) is the same, then I need only the first row of the max date available.
I'm really struguling as I'm not sure how to implement the logic into my code:
Select
table.person,
table.quantity
From
(
Select
table.date,
table.person,
table.country,
table.quantity,
ROW_NUMBER () over (
PARTITION by table.code, table.person
ORDER by table.date DESC
) AS rn
FROM
table
WHERE table.date >= DATE '{2020-04-10}' -5
) a
WHERE a.RN IN (1,2)
Is it possible to create a rule to sum rows 1 and 2 when country is different (Artur case) and only return row number 1 when the country is the same for a name (Joseph case)?
Use dense_rank() or max() as a window function:
select person, sum(quantity)
from (select t.*,
max(date) over (partition by person) as max_date
from t
) t
where date = max_date
group by person;
EDIT:
Hmmm . . . I think you might want one row per country per person on the max date. If so:
select person, sum(quantity)
from (select t.*,
row_number() over (partition by person, country order by date desc) as seqnum_pc,
rank() over (partition by person order by date desc) as seqnum_p
from t
) t
where seqnum_p = 1 and seqnum_pc = 1
group by person;
I'm writing a query to find duplicate records. I have table with following columns
Id, Deliveries, TankId, Timestamp.
I have inserted duplicate records, that is for same tankid, same deliveries with the +1 day offset timestamp.
Now I want to remove duplicate records which is with lesser timestamp.
e.g. I have duplicate deliveries added for same tankid on 24th and 25th july. I need to remove 24th record.
I tried the following query;
SELECT raw.TimeStamp,raw.[Delivery],raw.[TankId]
FROM [dbo].[tObservationData] raw
INNER JOIN (
SELECT [Delivery],[TankSystemId]
FROM [dbo].[ObservationData]
GROUP BY [Delivery],[TankSystemId]
HAVING COUNT([ObservationDataId]) > 1
) dup
ON raw.[Delivery] = dup.[Delivery] AND raw.[TankId] = dup.[TankId]
AND raw.TimeStamp >'2019-06-30 00:00:00.0000000' AND raw.[DeliveryL]>0
ORDER BY [TankSystemId],TimeStamp
But above gives other records too, how can I find and delete those duplicate records?
In this case you can use partition by order by clause. You can partition by TankID and Delivery and order by Timestamp in desc order
Select * from (
Select *,ROW_NUMBER() OVER (PARTITION BY TankID,Delievry ORDER BY [Timestamp] DESC) AS rn
from [dbo].[ObservationData]
)
where rn = 1
In the above code records with rn=1 will have the latest timestamp. So you can only select those and ignore others. Also you can use the same to remove/delete the records from you table.
WITH TempObservationdata (TankID,Delivery,Timestamp)
AS
(
SELECT TankID,Delivery,ROW_NUMBER() OVER(PARTITION by TankID, Delivery ORDER BY Timsetamp desc)
AS Timestamp
FROM dbo.ObservationData
)
--Now Delete Duplicate Rows
DELETE FROM TempObservationdata
WHERE Timestamp > 1
think it will work
SELECT raw.TimeStamp,raw.[Delivery],raw.[TankId]
FROM [dbo].[tObservationData] raw
INNER JOIN (
SELECT [Delivery],[TankSystemId],min([TimeStamp]) as min_ts
FROM [dbo].[ObservationData]
GROUP BY [Delivery],[TankSystemId]
HAVING COUNT([ObservationDataId]) > 1
) dup
ON raw.[Delivery] = dup.[Delivery] AND raw.[TankId] = dup.[TankId] and raw.[TimeStamp] = dup.min_ts
AND raw.TimeStamp >'2019-06-30 00:00:00.0000000' AND raw.[DeliveryL]>0
ORDER BY [TankSystemId],TimeStamp
Are you just looking for this?
SELECT od.*
FROM (SELECT od.*,
ROW_NUMBER() OVER (PARTITION BY od.TankId, od.Delivery ORDER BY od.TimeStamp DESC) as seqnum
FROM [dbo].[tObservationData] od
) od
WHERE seqnum = 1;
I've a situation where there is one ticket history table. it saves all the actions done against a ticket. how to write a query which will return the first record and the last record against specific ticket.
for example in the above table I've one ticket with id 78580. I want to get the first row and last row based on date column.
Just use row_number():
select t.*
from (select t.*,
row_number() over (partition by ticket_id order by action_when asc) as seqnum_a,
row_number() over (partition by ticket_id order by action_when desc) as seqnum_d
from tickets t
) t
where seqnum_a = 1 or seqnum_d = 1;
Use min and max to get first and last date, grouped by ticket id.
SELECT ticket_id, min(action_when), max(action_when)
FROM table_name
GROUP BY ticket_id;